How Does Dropping Rocks from Different Heights Affect Kinetic Energy Ratios?

AI Thread Summary
The discussion centers on calculating the kinetic energy ratios of rocks dropped from different heights above Earth's surface. When a rock is dropped from a height of RE, its kinetic energy K1 is derived from gravitational potential energy, while K2 for a rock dropped from 2RE is calculated similarly. The initial calculations suggested K2/K1 equals 2/3, but this was incorrect; the correct ratio is 4/3. The error arose from not considering the correct gravitational potential energy formula, which varies with distance. The proper approach involves using PEgrav = -GMm/r to accurately compare the changes in potential energy.
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Homework Statement


A rock is dropped from a distance RE above the surface of the Earth, and is observed to have a kinetic energy of K1, when it hits the ground. An identical rock is dropped from 2RE above the surface and has kinetic energy K2 when it hits the ground. Re is the radius of the earth. What is K2/K1.

a) 2 b) 4/3 c) 3/2 d) 2/3 e)4

Homework Equations

The Attempt at a Solution



K1= (-GMm)/(RE+ RE)= (−GMm)/2RE K2 = (-GMm)/(RE+ 2RE)= (−GMm)/3RE

K2/K1 = [(−GMm)/3RE]/[(−GMm)/2RE] = [(−GMm)/3RE] * [(2RE/(−GMm)] = 2/3 = d).
However the answer is 4/3, where did I go wrong?



 
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You want to compare the changes in PE that each rock undergoes when they fall. Where does each rock end up at then end of its fall? Not at infinity...
 
Oh I see where I made a mistake, thank you!
 
But how do we compare the changes in PE do we do mgRe?
 
Sparkling Eyes said:
But how do we compare the changes in PE do we do mgRe?
No. Those distances are too great to be taking g as a constant. Use PEgrav=-GMm/r, as in the OP.
 
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